Chem 1212k Lab Report

Topics: Water, Mole, Kilogram, Mass, Ammonia, Hydrochloric acid / Pages: 6 (1333 words) / Published: Mar 10th, 2013
Deborah Bell
April 17, 2012
Chemistry 1212K Lab Synthesis Report

Introduction In this Chemistry Lab the main objective is to perform accurate chemical analysis for the quantity of elements and compounds in a sample. There will be a compound made then synthesized. The methods used were acid-base titrations, redox titrations, gravity filtration, and distillation. General conclusions included
Weight of Crucibles

1. The first experiment is Preparation of a Cobalt Amine Bromide Product ; Synthesis #3 was used to create the compound. Added 5 grams of cobalt carbonate to 20 mL of hrdrobromic acid in a beaker. Noticied a slight color change to dark purple. Solution frothed after it settled I mixed in 15mL water and did a gravity filtration. Added the filtrate to a mixture of 2 grams activated charcoal and 25 mL of aqueous ammonia concentrated. Add 6mL of 30% hydrogen peroxide 3-4 drops at a time and heated. Transfered to a beaker and add a boiling solution of 3 mL HBr and 135 mL water.
Synthesis #3 yielded an orange colored powder.

2. Procedure for Analysis for percent Halide in a synthesized cobalt compound.
Prepare AgNO3 first: Weigh out 1.7-1.8 grams of AgNO3 and dissolve it in 50 mL deionized water. Mix in 10 drops of concentrated HNO3. Crucibles were already at a constant weight. Weighed out.24-.25 gram samples. Added 125-150 mL of deionized water and 10 drops of HNO3. Add 25 mL of AgNO3 solution to each sample. Boil gently. Collect AgX in gooch crucible. Used suction flask to dry then they were cooled and weighed.

My results yielded 50.1 % Halide ± .12
% Yield Calculation:
Crucible 1 weight : 31.7986 g
31.7986-31.2969 = .5017 g 35.452143.327X .5017= .1240
.1240.2478= .5004 x 100=50.04% Halide
Crucible 2 weight : 31.2134 g
31.2134 – 30.7106 = .5028
35.452143.327 X .5028= .1243
.1243.2478= .5016 x 100=50.16 % Halide

3. Procedure for Preparation and

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